Recent content by Masaki

Hello, tiny-tim, Actually, there is a problem just before this one, which requests to show (a) \displaystyle e^x > \sum_{k=0}^{n} \frac{x^k}{k!} and (b) \displaystyle \sum_{k=0}^{2m-1} \frac{(-1)^k x^k}{k!} < e^{-x} < \sum_{k=0}^{2m} \frac{(-1)^k x^k}{k!}. I could prove the first part by...

Homework Statement If n is a positive integer and if x > 0, show that \displaystyle\left(1 + \frac{x}{n}\right)^n < e^x and that \displaystyle e^x < \left(1 - \frac{x}{n}\right)^{-n} if \displaystyle x < n. The Attempt at a Solution I have proved the first inequality, but I am confused...

Thanks for the reply! I think the problem now becomes manageable. Let a_n = \arctan(1/\log{n}) and b_n = 1/\log{n}. Then, we have, with using l'Hôpital's rule, \displaystyle \lim_{n\rightarrow\infty} \frac{a_n}{b_n} = \lim_{n\rightarrow\infty} \frac{a_n'}{b_n'} = \lim_{n\rightarrow\infty}...

Homework Statement Determine (absolute/conditional) convergence or divergence of the series \displaystyle\sum_{n=1}^{\infty} (-1)^n \left(\frac{\pi}{2}-\text{arctan}({\log{n}})\right) The Attempt at a Solution It is easy to see that the series is (conditionally) convergent by Leibniz's...

Aha! It makes sense now! \displaystyle f(x) = \int_1^x \frac{\ln{t}}{t+1} dt = \int_1^{1/x} \frac{-\ln{y}}{\frac{1}{y}+1} \cdot \frac{-1}{y^2} dy = \int_1^{1/x} \frac{\ln{y}}{y(y+1)} dy = - f\left(\frac{1}{x}\right) + \int_1^{1/x} \frac{\ln{y}}{y} dy which gives the answer. Wow, I learned a...

Would you suggest how did you obtain an alternate representation for f(1/x) that can be added to f(x)? Just substituting u=xt into f(1/x) yields \displaystyle f\left(\frac{1}{x}\right) = \int_1^{1/x} \frac{\ln{t}}{t+1} dt = \ln{x} \int_1^x \frac{dt}{t+x} - \int_1^x \frac{\ln{t}}{t+x}dt and...

klondike, I got it, thanks! I've realized that forgetting to multiply by -1/x2 led me to the wrong way, and made me believe that FTC won't make it. Now I can have sweet dreams.:smile: SammyS, Apostol's Calculus has answers to numerical exercises in the back.

Homework Statement Let f(x)=∫1x (logt)/(t+1)dt if x>0. Compute f(x) + f(1/x). As a check, you should obtain f(2) + f(1/2) = 1/2(log2)2. The Attempt at a Solution I have tried various calculation that I know, such as integration by parts, but after ten hours with pieces of papers, I still...